# Negative Exponents in Binomial Theorem

I'm looking at extensions of the binomial formula to negative powers. I've figured out how to do $n \choose k$ when $n < 0$ and $k \geq 0$: $${n \choose k} = (-1)^k {-n + k - 1 \choose k}$$

So now let's look at one case for using the binomial coefficient: $$(1+x)^n = \sum_{k=0}^n {n \choose k}x^k$$

How do I evaluate $\sum_{k = 0}^{n}$ when $n < 0$? From searching around on the internet I think it's just an infinite series, i.e. $k$ keeps incrementing by 1 forever. But that gets me confused about

\begin{align*} (a + b)^n &= a^n(1 + \frac{b}{a})^n \\ &= a^n \left(\sum_{k = 0}^{n}{n \choose k}\left(\frac{b}{a}\right)^k\right)\\ &= a^n \left(1 + n \left(\frac{b}{a}\right) + \frac{(n)(n-1)}{2}\left(\frac{b}{a}\right)^2 + \cdots\right) \end{align*} and \begin{align*} (b + a)^n &= b^n\left(1 + \frac{a}{b}\right)^n\\ &= b^n \left(\sum_{k = 0}^{n}{n \choose k}\left(\frac{a}{b}\right)^k\right)\\ &= b^n \left(1 + n \left(\frac{a}{b}\right) + \frac{(n)(n-1)}{2}\left(\frac{a}{b}\right)^2 + \cdots\right) \end{align*}

Now the two should be equal, but in the first sum I'd never get a $b^n$ and in the second sum I'd never get a $a^n$?

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The first sum only converges if $\left| \frac{a}{b} \right| > 1$ whereas the second sum only converges if this is less than $1$, so the two series are never simultaneously valid. – Qiaochu Yuan Nov 26 '11 at 4:36
Ok, so my interpretation of $\sum_{k = 0}^{n}$ is correct for $n < 0$? I see that you are correct. Does this mean that the formula $(1+x)^n = \sum_{k=0}^n {n \choose k}x^k$ is valid for all x is $n \geq 0$ and only for $|x| < 1$ when $n < 0$? I have not seen this condition before for this formula. – Tianxiang Xiong Nov 26 '11 at 4:43
Correct. In fact $n$ may be any complex number and the sum will be infinite unless $n$ is a non-negative integer. – Qiaochu Yuan Nov 26 '11 at 4:47
Ok. I suppose that this should also be extended to $(x+y)^n$ as well, where it equals $\sum_{k=0}^n {n \choose k}x^{n-k}y^k$ iff $|\frac{y}{x}| < 1$ and $\sum_{k=0}^n {n \choose k}x^{k}y^{n-k}$ iff $|\frac{x}{y}| < 1$? Thanks for your help. – Tianxiang Xiong Nov 26 '11 at 4:55
Your sums should go from 0 to $\infty$, not 0 to $n$. – Ted Nov 26 '11 at 5:51

The below is too long for a comment so I'm including it here even though I'm not sure it "answers" the question.

If you think about $(1+x)^{-n}$ as living in the ring of formal power series $\mathbb{Z}[[x]]$, then you can show that $$(1+x)^{-n} = \sum_{k=0}^{\infty} (-1)^k \binom{n+k-1}{k} x^k$$ and the identity $\binom{-n}{k} = (-1)^k \binom{n+k-1}{k}$ seems very natural. Here's how...

First expand $(1+x)^{-n} = \bigg(\frac{1}{1-(-x)}\bigg)^n = (1 - x + x^2 - x^3 + \dots)^n$. Now, the coefficient on $x^k$ in that product is simply the number of ways to write $k$ as a sum of $n$ nonnegative numbers. That set of sums is in bijection to the set of diagrams with $k$ stars with $n-1$ bars among them. (For example, suppose $k=9$ and $n=4$. Then, **|*|***|*** corresponds to the sum $9=2+1+3+3$; ****||***|**corresponds to the sum $9 = 4+0+3+2$; ****|***||** corresponds to $9=4+3+0+2$; etc.) In each of these stars-and-bars diagrams we have $n+k-1$ objects, and we choose which ones are the $k$ stars in $\binom{n+k-1}{k}$ many ways. The $(-1)^k$ term comes from the alternating signs, and that proves the sum.

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This is exactly the sort of explanation I was looking for. I wanted to find one in terms of stars and bars: you rock. Also, the phrase "bijection to the set of diagrams with stars and bars" is epic. I want to say that all day long. I was wondering, though (and I guess this is another question entirely): why is (1 + x)^-n the number of ways to write k as a sum of n integers? What happens to the negative terms? – Ziggy Oct 28 '12 at 22:52
How do you know that the coefficient is indeed the number of n-tuples of non-negative integers whose sum is k? Can you give more details in this, please? – Ana Dec 2 '14 at 2:41

For a=1, the negative binomial series simplifies to (x+1)^(-n)=1-nx+1/2n(n+1)x^2-1/6n(n+1)(n+2)x^3+....

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